The battery in the diagram is to be charged by the generator $G$. The generator has a terminal voltage of $120$ $\mathrm{volts}$ when the charging current is $10$ $\mathrm{amperes}.$ The battery has an $\mathrm{emf}$ of $100$ $\mathrm{volts}$ and an internal resistance of $1$ $\mathrm{ohm}.$ In order to charge the battery at $10$ $\mathrm{amperes}$ charging current, the resistance $R$ should be set at ................ $\Omega$
A$0.1$
B$0.5$
C$1$
D$5$
Medium
Download our app for free and get started
C$1$
c The generator has a terminal voltage of $120$ volts when the charging current is $10$ amperes. Hence,
$E=V+I r=120+10(1)=130 \mathrm{V}$
Hence, in circuit
$R=\frac{E-V}{I}$
$R=\frac{130-120}{10}=1 \Omega$
Download our app
and get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Consider the circuit shown below where all resistors are $1 \,k \Omega$. If a current of magnitude $1 \,mA$ flows through the resistor marked $X$, the potential difference measured between points $P$ and $Q$ are ..............$V$
A $6.0\,volt$ battery is connected to two light bulbs as shown in figure. Light bulb $1$ has resistance $3\,ohm$ while light bulb $2$ has resistance $6\,ohm.$ Battery has negligible internal resistance. Which bulb will glow brighter?
Consider a wire having current $10\,A$ having area of crossection $1\,cm^2$. If number of electrons per unit volume is $9 \times 10^{28}\, m^{-3}$. Find the drift velocity of electrons
The resistance of a wire is $20\, ohms$. It is so stretched that the length becomes three times, then the new resistance of the wire will be ............. $ohms$
There are $n$ similar conductors each of resistance $R$. The resultant resistance comes out to be $x$ when connected in parallel. If they are connected in series, the resistance comes out to be
In an experiment of potentiometer for measuring the internal resistance of primary cell a balancing length $\ell $ is obtained on the potentiometer wire when the cell is open circuit. Now the cell is short circuited by a resistance $R$. If $R$ is to be equal to the internal resistance of the cell the balancing length on the potentiometer wire will be